Question Number 7360 by rohit meena last updated on 24/Aug/16
Answered by Rasheed Soomro last updated on 25/Aug/16
![(3) (4−k)x^2 +2(k+2)x+8k+1 The value of k,for which the above expression be perfect square, is required. −−−−−−−−−−−−−−−−−−−−−−−− A quadratic expression will be perfect square if it is of the following form: a^2 ±2ab+b^2 If first term, middle term and last term are f, m and l respectively then m=±2(√f) (√l) Dividing by 4−k to the given expression x^2 +((2(k+2))/(4−k))x+((8k+1)/(4−k)) [To make the first term perfect square] x^2 +((2(k+2))/(4−k))x+((√((8k+1)/(4−k))))^2 [Writing last term in square form] Now , m=±2(√f) (√l) ((2(k+2))/(4−k))x=±2(x)((√((8k+1)/(4−k)))) (((k+2))/(4−k))=±(√((8k+1)/(4−k))) (((k+2)/(4−k)))^2 =(±(√((8k+1)/(4−k))))^2 (((k+2)^2 )/((4−k)^2 ))=((8k+1)/(4−k)) ((k^2 +4k+4)/(4−k))=8k+1 k^2 +4k+4=32k+4−8k^2 −k 9k^2 −27k=0 k(k−3)=0 k=0 ∣ k=3 (4) x−α is a common factor of a_1 x^2 +b_1 x+c and a_2 x^2 +b_2 x+c x−k is a factor of p(x)⇔k is a root of p(x)=0 x−α is a common factor of a_1 x^2 +b_1 x+c and a_2 x^2 +b_2 x+c ∴ α is a root of both equations: a_1 x^2 +b_1 x+c=0 and a_2 x^2 +b_2 x+c=0 Hence, a_1 α^2 +b_1 α+c=0 ∧ a_2 α^2 +b_2 α+c=0 On subtracting: (a_1 −a_2 )α^2 +(b_1 −b_2 )α=0 (a_1 −a_2 )α+(b_1 −b_2 )=0 (a_1 −a_2 )α=−(b_1 −b_2 ) (a_1 −a_2 )α=b_2 −b_1 P r o v e d](https://www.tinkutara.com/question/Q7364.png)
$$\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{4}−{k}\right){x}^{\mathrm{2}} +\mathrm{2}\left({k}+\mathrm{2}\right){x}+\mathrm{8}{k}+\mathrm{1} \\ $$$${The}\:{value}\:{of}\:\:{k},{for}\:{which}\:{the}\:{above} \\ $$$${expression}\:{be}\:{perfect}\:{square},\:{is}\:{required}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${A}\:{quadratic}\:{expression}\:{will}\:{be}\:{perfect}\:{square} \\ $$$${if}\:{it}\:{is}\:{of}\:{the}\:{following}\:{form}: \\ $$$${a}^{\mathrm{2}} \pm\mathrm{2}{ab}+{b}^{\mathrm{2}} \\ $$$${If}\:\:{first}\:{term},\:{middle}\:{term}\:{and}\:{last}\:{term}\:{are}\: \\ $$$${f},\:{m}\:\:{and}\:\:{l}\:{respectively}\:{then} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{m}=\pm\mathrm{2}\sqrt{{f}}\:\sqrt{{l}} \\ $$$${Dividing}\:{by}\:\mathrm{4}−{k}\:{to}\:{the}\:{given}\:{expression} \\ $$$$\:\:\:\:\:\:\:{x}^{\mathrm{2}} +\frac{\mathrm{2}\left({k}+\mathrm{2}\right)}{\mathrm{4}−{k}}{x}+\frac{\mathrm{8}{k}+\mathrm{1}}{\mathrm{4}−{k}}\:\:\left[{To}\:{make}\:{the}\:{first}\:{term}\:{perfect}\:{square}\right] \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +\frac{\mathrm{2}\left({k}+\mathrm{2}\right)}{\mathrm{4}−{k}}{x}+\left(\sqrt{\frac{\mathrm{8}{k}+\mathrm{1}}{\mathrm{4}−{k}}}\right)^{\mathrm{2}} \left[{Writing}\:{last}\:{term}\:{in}\:{square}\:{form}\right] \\ $$$${Now}\:,\:\:\:\:\:\:\:\:\:\:\:\:\:{m}=\pm\mathrm{2}\sqrt{{f}}\:\sqrt{{l}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}\left({k}+\mathrm{2}\right)}{\mathrm{4}−{k}}{x}=\pm\mathrm{2}\left({x}\right)\left(\sqrt{\frac{\mathrm{8}{k}+\mathrm{1}}{\mathrm{4}−{k}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\left({k}+\mathrm{2}\right)}{\mathrm{4}−{k}}=\pm\sqrt{\frac{\mathrm{8}{k}+\mathrm{1}}{\mathrm{4}−{k}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{{k}+\mathrm{2}}{\mathrm{4}−{k}}\right)^{\mathrm{2}} =\left(\pm\sqrt{\frac{\mathrm{8}{k}+\mathrm{1}}{\mathrm{4}−{k}}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\left({k}+\mathrm{2}\right)^{\mathrm{2}} }{\left(\mathrm{4}−{k}\right)^{\mathrm{2}} }=\frac{\mathrm{8}{k}+\mathrm{1}}{\mathrm{4}−{k}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{4}}{\mathrm{4}−{k}}=\mathrm{8}{k}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{4}=\mathrm{32}{k}+\mathrm{4}−\mathrm{8}{k}^{\mathrm{2}} −{k} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{9}{k}^{\mathrm{2}} −\mathrm{27}{k}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{k}\left({k}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{k}=\mathrm{0}\:\:\:\mid\:\:\:{k}=\mathrm{3} \\ $$$$\left(\mathrm{4}\right) \\ $$$${x}−\alpha\:{is}\:{a}\:{common}\:{factor}\:{of}\:{a}_{\mathrm{1}} {x}^{\mathrm{2}} +{b}_{\mathrm{1}} {x}+{c}\:\:{and} \\ $$$${a}_{\mathrm{2}} {x}^{\mathrm{2}} +{b}_{\mathrm{2}} {x}+{c} \\ $$$${x}−{k}\:{is}\:{a}\:{factor}\:{of}\:{p}\left({x}\right)\Leftrightarrow{k}\:\:{is}\:\:{a}\:{root}\:{of}\:\:{p}\left({x}\right)=\mathrm{0} \\ $$$${x}−\alpha\:{is}\:{a}\:{common}\:{factor}\:{of}\:{a}_{\mathrm{1}} {x}^{\mathrm{2}} +{b}_{\mathrm{1}} {x}+{c}\:\:{and}\:\:{a}_{\mathrm{2}} {x}^{\mathrm{2}} +{b}_{\mathrm{2}} {x}+{c} \\ $$$$\therefore\:\alpha\:{is}\:{a}\:{root}\:{of}\:{both}\:{equations}:\:\:{a}_{\mathrm{1}} {x}^{\mathrm{2}} +{b}_{\mathrm{1}} {x}+{c}=\mathrm{0} \\ $$$$\:\:\:\:\:{and}\:{a}_{\mathrm{2}} {x}^{\mathrm{2}} +{b}_{\mathrm{2}} {x}+{c}=\mathrm{0} \\ $$$${Hence},\:\:\:{a}_{\mathrm{1}} \alpha^{\mathrm{2}} +{b}_{\mathrm{1}} \alpha+{c}=\mathrm{0}\:\:\wedge\:\:\:{a}_{\mathrm{2}} \alpha^{\mathrm{2}} +{b}_{\mathrm{2}} \alpha+{c}=\mathrm{0} \\ $$$${On}\:{subtracting}:\:\:\left({a}_{\mathrm{1}} −{a}_{\mathrm{2}} \right)\alpha^{\mathrm{2}} +\left({b}_{\mathrm{1}} −{b}_{\mathrm{2}} \right)\alpha=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a}_{\mathrm{1}} −{a}_{\mathrm{2}} \right)\alpha+\left({b}_{\mathrm{1}} −{b}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a}_{\mathrm{1}} −{a}_{\mathrm{2}} \right)\alpha=−\left({b}_{\mathrm{1}} −{b}_{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a}_{\mathrm{1}} −{a}_{\mathrm{2}} \right)\alpha={b}_{\mathrm{2}} −{b}_{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{P}\:\:{r}\:\:{o}\:\:{v}\:\:{e}\:\:{d} \\ $$
Commented by rohit meena last updated on 25/Aug/16
$${or}\:{Qvestion}\:{ke}\:{answer} \\ $$$$ \\ $$
Commented by rohit meena last updated on 25/Aug/16
$${or}\:{Qvestion}\:{ke}\:{answer} \\ $$$${c}\:{kaha}\:{gya}\:{what}\:{value}\:{of}\:\boldsymbol{\mathrm{c}} \\ $$
Commented by Rasheed Soomro last updated on 26/Aug/16
$$\:\:\:\:\:\:\:{a}_{\mathrm{1}} \alpha^{\mathrm{2}} +{b}_{\mathrm{1}} \alpha+{c}=\mathrm{0} \\ $$$$\:\:\:\:\:_{−} {a}_{\mathrm{2}} \alpha^{\mathrm{2}} \underset{−} {+}{b}_{\mathrm{2}} \alpha\underset{−} {+}{c}=\mathrm{0} \\ $$$$−−−−−−−−−−−−− \\ $$$$\:\:\:\:\:\:\left({a}_{\mathrm{1}} −{a}_{\mathrm{2}} \right)\alpha^{\mathrm{2}} +\left({b}_{\mathrm{1}} −{b}_{\mathrm{2}} \right)\alpha+\mathrm{0}=\mathrm{0} \\ $$
Answered by sandy_suhendra last updated on 25/Aug/16
![3) perfect square ⇒ x_1 =x_2 ⇒ D=b^2 −4ac=0 a=4−k ; b=2(k+2) ; c=8k+1 D=[2(k+2)]^2 −4(4−k)(8k+1)=0 4k^2 +16k+16+32k^2 −124k−16=0 36k^2 −108k=0 36k(k−3)=0 k=0 or k=3 4) Mr Soomro has given the great answer continued](https://www.tinkutara.com/question/Q7367.png)
$$\left.\mathrm{3}\right)\:{perfect}\:{square}\:\Rightarrow\:{x}_{\mathrm{1}} ={x}_{\mathrm{2}} \:\Rightarrow\:{D}={b}^{\mathrm{2}} −\mathrm{4}{ac}=\mathrm{0} \\ $$$${a}=\mathrm{4}−{k}\:;\:{b}=\mathrm{2}\left({k}+\mathrm{2}\right)\:;\:{c}=\mathrm{8}{k}+\mathrm{1} \\ $$$${D}=\left[\mathrm{2}\left({k}+\mathrm{2}\right)\right]^{\mathrm{2}} −\mathrm{4}\left(\mathrm{4}−{k}\right)\left(\mathrm{8}{k}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{4}{k}^{\mathrm{2}} +\mathrm{16}{k}+\mathrm{16}+\mathrm{32}{k}^{\mathrm{2}} −\mathrm{124}{k}−\mathrm{16}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{36}{k}^{\mathrm{2}} −\mathrm{108}{k}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{36}{k}\left({k}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{k}=\mathrm{0}\:{or}\:{k}=\mathrm{3} \\ $$$$ \\ $$$$\left.\mathrm{4}\right)\:{Mr}\:{Soomro}\:{has}\:{given}\:{the}\:{great}\:{answer} \\ $$$$ \\ $$$${continued} \\ $$
Answered by sandy_suhendra last updated on 25/Aug/16
![6) a=4 ; b=−4(α−2) ; c=α−2 i) real n distinct ⇒ D>0 [−4(α−2)]^2 −4.4(α−2)>0 16α^2 −64α+64−16α+32>0 16α^2 −80α+96>0 α^2 −5α+6>0 (α−2)(α−3)>0 α<2 or α>3 ii)equal ⇒D=0 (α−2)(α−3)=0 α=2 or α=3 iii)imaginary ⇒D<0 (α−2)(α−3)<0 2<α<3 iv) opposite in sign ⇒ x_1 .x_2 <0 (c/a)<0 ((α−2)/4)<0 α−2<0 α<2 v)equal in magnitude but opposite sign ⇒ x_1 +x_2 =0 ((−b)/a)=0 b=0 −4(α−2)=0 α=2 continued](https://www.tinkutara.com/question/Q7368.png)
$$\left.\mathrm{6}\right)\:{a}=\mathrm{4}\:;\:{b}=−\mathrm{4}\left(\alpha−\mathrm{2}\right)\:;\:{c}=\alpha−\mathrm{2} \\ $$$$\left.{i}\right)\:{real}\:{n}\:{distinct}\:\Rightarrow\:{D}>\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[−\mathrm{4}\left(\alpha−\mathrm{2}\right)\right]^{\mathrm{2}} −\mathrm{4}.\mathrm{4}\left(\alpha−\mathrm{2}\right)>\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{16}\alpha^{\mathrm{2}} −\mathrm{64}\alpha+\mathrm{64}−\mathrm{16}\alpha+\mathrm{32}>\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{16}\alpha^{\mathrm{2}} −\mathrm{80}\alpha+\mathrm{96}>\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha^{\mathrm{2}} −\mathrm{5}\alpha+\mathrm{6}>\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\alpha−\mathrm{2}\right)\left(\alpha−\mathrm{3}\right)>\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha<\mathrm{2}\:{or}\:\alpha>\mathrm{3} \\ $$$$\left.{ii}\right){equal}\:\Rightarrow{D}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\alpha−\mathrm{2}\right)\left(\alpha−\mathrm{3}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha=\mathrm{2}\:{or}\:\alpha=\mathrm{3} \\ $$$$\left.{iii}\right){imaginary}\:\Rightarrow{D}<\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\alpha−\mathrm{2}\right)\left(\alpha−\mathrm{3}\right)<\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}<\alpha<\mathrm{3} \\ $$$$\left.{iv}\right)\:{opposite}\:{in}\:{sign}\:\Rightarrow\:{x}_{\mathrm{1}} .{x}_{\mathrm{2}} <\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{c}}{{a}}<\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\alpha−\mathrm{2}}{\mathrm{4}}<\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha−\mathrm{2}<\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha<\mathrm{2} \\ $$$$\left.{v}\right){equal}\:{in}\:{magnitude}\:{but}\:{opposite}\:{sign}\:\Rightarrow\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{−{b}}{{a}}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{4}\left(\alpha−\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha=\mathrm{2} \\ $$$${continued} \\ $$
Commented by sandy_suhendra last updated on 25/Aug/16
$$\alpha<\mathrm{2}\:{it}\:{means}\:−\infty<\alpha<\mathrm{2}\:\Rightarrow\left(−\infty,\mathrm{2}\right) \\ $$$$\alpha>\mathrm{3}\:{it}\:{means}\:\mathrm{3}<\alpha<\infty\:\Rightarrow\left(\mathrm{3},\infty\right) \\ $$
Commented by rohit meena last updated on 25/Aug/16
$${answer}\:{is}\:\left({i}\right)\:\left(−\infty,\mathrm{2}\right)\:\left(\mathrm{3},\infty\right) \\ $$
Answered by sandy_suhendra last updated on 25/Aug/16
$$\left.\mathrm{7}\right)\:{P}\left({x}\right).{Q}\left({x}\right)=\mathrm{0} \\ $$$$\left({ax}^{\mathrm{2}} +{bx}+{c}\right)\left(−{ax}^{\mathrm{2}} +{dx}+{c}\right)=\mathrm{0} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:\Rightarrow\:{the}\:{roots}\:{x}_{\mathrm{1}} \:{and}\:{x}_{\mathrm{2}} \\ $$$${or}\:−{ax}^{\mathrm{2}} +{dx}+{c}=\mathrm{0}\:\Rightarrow\:{the}\:{roots}\:{x}_{\mathrm{3}} \:{and}\:{x}_{\mathrm{4}} \\ $$$${x}_{\mathrm{1},\mathrm{2}} \:=\:\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\: \\ $$$${x}_{\mathrm{3},\mathrm{4}} \:=\:\frac{−{d}\pm\sqrt{{d}^{\mathrm{2}} −\mathrm{4}\left(−{a}\right){c}}}{−\mathrm{2}{a}}\:=\:\frac{−{d}\pm\sqrt{{d}^{\mathrm{2}} +\mathrm{4}{ac}}}{−\mathrm{2}{a}}\: \\ $$$${if}\:{ac}>\mathrm{0}\:\wedge\:{b}^{\mathrm{2}} −\mathrm{4}{ac}<\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{1},\mathrm{2}} \:{are}\:{imaginary}\:{and}\:{x}_{\mathrm{3},\mathrm{4}} \:{are}\:{real}\:\left(\mathrm{2}\:{real}\:{roots}\right) \\ $$$${if}\:{ac}>\mathrm{0}\:\wedge\:{b}^{\mathrm{2}} −\mathrm{4}{ac}>\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{1},\mathrm{2}} \:{and}\:{x}_{\mathrm{3},\mathrm{4}} \:{are}\:{real}\:\left(\mathrm{4}\:{real}\:{roots}\right) \\ $$$${if}\:{ac}<\mathrm{0}\:\wedge\:{d}^{\mathrm{2}} +\mathrm{4}{ac}>\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{1},\mathrm{2}} \:{and}\:{x}_{\mathrm{3},\mathrm{4}} \:{are}\:{real}\:\left(\mathrm{4}\:{real}\:{roots}\right) \\ $$$${if}\:{ac}<\mathrm{0}\:\wedge\:{d}^{\mathrm{2}} +\mathrm{4}{ac}<\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{1},\mathrm{2}} \:{are}\:{real}\:{and}\:{x}_{\mathrm{3},\mathrm{4}} \:{are}\:{imaginary}\:\left(\mathrm{2}\:{real}\:{roots}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 25/Aug/16
$${x}_{\mathrm{3},\mathrm{4}} \:=\:\frac{−{d}\pm\sqrt{{d}^{\mathrm{2}} −\mathrm{4}\left(−{a}\right){c}}}{−\mathrm{2}{a}} \\ $$
Commented by sandy_suhendra last updated on 25/Aug/16
$${Thank}'{s}\:{Rasheed}\:{for}\:{your}\:{correction}.\:{I}'{ve}\:{corrected}\:{it} \\ $$
Commented by Rasheed Soomro last updated on 25/Aug/16
$${Why}\:{d}^{\mathrm{2}} +\mathrm{4}{ac}\:{is}\:{always}\:{positive}?\:{Is}\:\:{there}\:{a}\:{guarantee} \\ $$$${that}\:\:\:\:\:\:\:\:\:\:\:{a}>\mathrm{0}\:\wedge\:{c}>\mathrm{0}\:\:\:\:\:{or}\:\:\:\:{a}<\mathrm{0}\:\wedge\:{c}<\mathrm{0} \\ $$$${However}\:{we}\:{can}\:\:{say}\:{that}\:\boldsymbol{{at}}\:\boldsymbol{{least}}\:\boldsymbol{{one}}\:\boldsymbol{{of}} \\ $$$${b}^{\mathrm{2}} −\mathrm{4}{ac}\:\:\boldsymbol{{and}}\:\:{d}^{\mathrm{2}} +\mathrm{4}{ac}\:\boldsymbol{{is}}\:\boldsymbol{{positive}}.{Because} \\ $$$${there}\:{are}\:{two}\:{possibilities}\:{ac}>\mathrm{0}\:{or}\:{ac}<\mathrm{0}. \\ $$$${If}\:{ac}>\mathrm{0}\:{then}\:{d}^{\mathrm{2}} +\mathrm{4}{ac}>\mathrm{0} \\ $$$${If}\:{ac}<\mathrm{0}\:{then}\:{b}^{\mathrm{2}} −\mathrm{4}{ac}>\mathrm{0} \\ $$$$\left({Third}\:{possibility}\:{ac}=\mathrm{0}\:{is}\:{excluded}\:{in}\:{the}\:{question}.\right) \\ $$
Commented by sandy_suhendra last updated on 25/Aug/16
$${That}'{s}\:{great},\:{you}'{re}\:{right}!\:{Thank}'{s}\:{Rasheed}\:{for}\:{reminding}\:{me}.\:{I}\:{have}\:{made}\:{a}\:{mistake}.\:{It}\:{has}\:{been}\:{corrected}. \\ $$
Commented by Rasheed Soomro last updated on 25/Aug/16
$${Your}\:{logic}\:{is}\:{simple}\:{and}\:{admirable}. \\ $$
Answered by Yozzia last updated on 25/Aug/16
![5) 3x^2 +2αxy+2y^2 +2ax−4y+1 can be resolved into two linear factors. ∴ 3x^2 +2αxy+2y^2 +2ax−4y+1≡(px+by+c)(dx+ey+f) −−−−−−−−−−−−−−−−−−−−−−−−−− y=0⇒3x^2 +2ax+1≡(px+c)(dx+f) 3x^2 +2(a/( (√3)))(√3)x+(a^2 /3)−(a^2 /3)+1≡(px+c)(dx+f) ((√3)x+(a/( (√3)))+(√((a^2 −3)/3)))((√3)x+(a/( (√3)))−(√((a^2 −3)/3)))≡(px+c)(dx+f) ⇒p=d=(√3), c=((a+(√(a^2 −3)))/( (√3))), f=((a−(√(a^2 −3)))/( (√3))) −−−−−−−−−−−−−−−−−−−−−−−−−−− x=0⇒2y^2 −4y+1≡(by+c)(ey+f) 2y^2 −2×(√2)(√2)y+2−2+1≡(by+c)(ey+f) ((√2)y−(√2))^2 −1≡(by+c)(ey+f) ((√2)y−(√2)−1)((√2)y−(√2)+1)≡(by+c)(ey+f) ⇒b=e=(√2) and possibly c=−(√2)−1 and f=1−(√2) or c=1−(√2) and f=−1−(√2). −−−−−−−−−−−−−−−−−−−−−−−− coefficient of xy=2α=pe+bd 2α=(√2)(p+d)=(√2)(2(√3))⇒α=(√6). −−−−−−−−−−−−−−−−−−−−−−− We have that x^2 +4ax+2a^2 +6=0. Try x=α=(√6). ∴ 6+4a(√6)+2a^2 +6=0 2a^2 +4a(√6)+12=0. Now, if c=−1−(√2)⇒a^2 −3=((−1−(√2))(√3)−a)^2 a^2 −3=a^2 +2a(√3)(1+(√2))+3(1+(√2))^2 2a(1+(√2))(√3)+3+3(3+2(√2))=0 2a(1+(√2))(√3)+12+6(√2)=0 a(1+(√2))×3+(6+3(√2))(√3)=0 a(1+(√2))+(2+(√2))(√3)=0 a=((−(2+(√2))(√3))/(1+(√2)))=(2+(√2))(1−(√2))(√3)=−(√6) Also, if f=1−(√2)⇒a=−(√6). [((a−(√(a^2 −3)))/( (√3)))=1−(√2)⇒(a−(1−(√2))(√3))^2 =a^2 −3 2a((√2)−1)(√3)+3(3−2(√2))=−3 a(−1+(√2))+(2−(√2))(√3)=0 a=(((√2)−2)/(−1+(√2)))(√3)=((((√2)−2)(−1−(√2)))/(1−2))(√3) a=(√3)((√2)−2)(1+(√2))=(√3)((√2)−2(√2)−2+2)=−(√6)] ∴2(−(√6))^2 +4(√6)(−(√6))+12=0 12+12−24=0 Hence α satisfies x^2 +4ax+2a^2 +6=0 for some constant a, if we can resolve 3x^2 +2αxy+2y^2 +2ax−4y+1 into two linear factors.](https://www.tinkutara.com/question/Q7373.png)
$$\left.\mathrm{5}\right)\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}\alpha{xy}+\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{ax}−\mathrm{4}{y}+\mathrm{1} \\ $$$${can}\:{be}\:{resolved}\:{into}\:{two}\:{linear}\:{factors}. \\ $$$$\therefore\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}\alpha{xy}+\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{ax}−\mathrm{4}{y}+\mathrm{1}\equiv\left({px}+{by}+{c}\right)\left({dx}+{ey}+{f}\right) \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${y}=\mathrm{0}\Rightarrow\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{ax}+\mathrm{1}\equiv\left({px}+{c}\right)\left({dx}+{f}\right) \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}\frac{{a}}{\:\sqrt{\mathrm{3}}}\sqrt{\mathrm{3}}{x}+\frac{{a}^{\mathrm{2}} }{\mathrm{3}}−\frac{{a}^{\mathrm{2}} }{\mathrm{3}}+\mathrm{1}\equiv\left({px}+{c}\right)\left({dx}+{f}\right) \\ $$$$\left(\sqrt{\mathrm{3}}{x}+\frac{{a}}{\:\sqrt{\mathrm{3}}}+\sqrt{\frac{{a}^{\mathrm{2}} −\mathrm{3}}{\mathrm{3}}}\right)\left(\sqrt{\mathrm{3}}{x}+\frac{{a}}{\:\sqrt{\mathrm{3}}}−\sqrt{\frac{{a}^{\mathrm{2}} −\mathrm{3}}{\mathrm{3}}}\right)\equiv\left({px}+{c}\right)\left({dx}+{f}\right) \\ $$$$\Rightarrow{p}={d}=\sqrt{\mathrm{3}},\:{c}=\frac{{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{3}}}{\:\sqrt{\mathrm{3}}},\:{f}=\frac{{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{3}}}{\:\sqrt{\mathrm{3}}} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${x}=\mathrm{0}\Rightarrow\mathrm{2}{y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{1}\equiv\left({by}+{c}\right)\left({ey}+{f}\right) \\ $$$$\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}×\sqrt{\mathrm{2}}\sqrt{\mathrm{2}}{y}+\mathrm{2}−\mathrm{2}+\mathrm{1}\equiv\left({by}+{c}\right)\left({ey}+{f}\right) \\ $$$$\left(\sqrt{\mathrm{2}}{y}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{1}\equiv\left({by}+{c}\right)\left({ey}+{f}\right) \\ $$$$\left(\sqrt{\mathrm{2}}{y}−\sqrt{\mathrm{2}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}{y}−\sqrt{\mathrm{2}}+\mathrm{1}\right)\equiv\left({by}+{c}\right)\left({ey}+{f}\right) \\ $$$$\Rightarrow{b}={e}=\sqrt{\mathrm{2}}\:{and}\:{possibly}\:{c}=−\sqrt{\mathrm{2}}−\mathrm{1}\:{and}\:{f}=\mathrm{1}−\sqrt{\mathrm{2}} \\ $$$${or}\:{c}=\mathrm{1}−\sqrt{\mathrm{2}}\:{and}\:{f}=−\mathrm{1}−\sqrt{\mathrm{2}}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${coefficient}\:{of}\:{xy}=\mathrm{2}\alpha={pe}+{bd} \\ $$$$\mathrm{2}\alpha=\sqrt{\mathrm{2}}\left({p}+{d}\right)=\sqrt{\mathrm{2}}\left(\mathrm{2}\sqrt{\mathrm{3}}\right)\Rightarrow\alpha=\sqrt{\mathrm{6}}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${We}\:{have}\:{that}\:{x}^{\mathrm{2}} +\mathrm{4}{ax}+\mathrm{2}{a}^{\mathrm{2}} +\mathrm{6}=\mathrm{0}. \\ $$$${Try}\:{x}=\alpha=\sqrt{\mathrm{6}}. \\ $$$$\therefore\:\mathrm{6}+\mathrm{4}{a}\sqrt{\mathrm{6}}+\mathrm{2}{a}^{\mathrm{2}} +\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{4}{a}\sqrt{\mathrm{6}}+\mathrm{12}=\mathrm{0}. \\ $$$${Now},\:{if}\:{c}=−\mathrm{1}−\sqrt{\mathrm{2}}\Rightarrow{a}^{\mathrm{2}} −\mathrm{3}=\left(\left(−\mathrm{1}−\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{3}}−{a}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} −\mathrm{3}={a}^{\mathrm{2}} +\mathrm{2}{a}\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)+\mathrm{3}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{a}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{3}}+\mathrm{3}+\mathrm{3}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\mathrm{2}{a}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{3}}+\mathrm{12}+\mathrm{6}\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$${a}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)×\mathrm{3}+\left(\mathrm{6}+\mathrm{3}\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$${a}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)+\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$${a}=\frac{−\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{3}}}{\mathrm{1}+\sqrt{\mathrm{2}}}=\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{3}}=−\sqrt{\mathrm{6}} \\ $$$$ \\ $$$${Also},\:{if}\:{f}=\mathrm{1}−\sqrt{\mathrm{2}}\Rightarrow{a}=−\sqrt{\mathrm{6}}. \\ $$$$\left[\frac{{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{3}}}{\:\sqrt{\mathrm{3}}}=\mathrm{1}−\sqrt{\mathrm{2}}\Rightarrow\left({a}−\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{3}}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} −\mathrm{3}\right. \\ $$$$\mathrm{2}{a}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\sqrt{\mathrm{3}}+\mathrm{3}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)=−\mathrm{3} \\ $$$${a}\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)+\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$${a}=\frac{\sqrt{\mathrm{2}}−\mathrm{2}}{−\mathrm{1}+\sqrt{\mathrm{2}}}\sqrt{\mathrm{3}}=\frac{\left(\sqrt{\mathrm{2}}−\mathrm{2}\right)\left(−\mathrm{1}−\sqrt{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{2}}\sqrt{\mathrm{3}} \\ $$$$\left.{a}=\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{2}}−\mathrm{2}\right)\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)=\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}+\mathrm{2}\right)=−\sqrt{\mathrm{6}}\right] \\ $$$$ \\ $$$$\therefore\mathrm{2}\left(−\sqrt{\mathrm{6}}\right)^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{6}}\left(−\sqrt{\mathrm{6}}\right)+\mathrm{12}=\mathrm{0} \\ $$$$\mathrm{12}+\mathrm{12}−\mathrm{24}=\mathrm{0} \\ $$$${Hence}\:\alpha\:{satisfies}\:{x}^{\mathrm{2}} +\mathrm{4}{ax}+\mathrm{2}{a}^{\mathrm{2}} +\mathrm{6}=\mathrm{0} \\ $$$${for}\:{some}\:{constant}\:{a},\:{if}\:{we}\:{can}\:{resolve} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}\alpha{xy}+\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{ax}−\mathrm{4}{y}+\mathrm{1}\:{into}\:{two} \\ $$$${linear}\:{factors}. \\ $$$$ \\ $$$$ \\ $$